alexmark Posted July 16, 2012 Share Posted July 16, 2012 Hi all I am working on a project and I can not seem to get this right no mater what I do. I have 3 tables in db, year, manufacturer_id and models I would like to display makes and models based on requests made by user. I.e., if a user selects Make as Ford and then Year as 1942 the Model should return a list of all models produced in that year. However it does not work can some one tell me where I am going wrong please. My data bases tables are set up as follows. manufacturer 2 columns id (which is unique) and manufacturer year 2 columns id (which is unique) and year model 4 columns id (which is unique) year (which corresponds to year id table) manufacturer_id (which corresponds to manufactures id table) and model. The code I am using is as follows. <tr> <td align="right" label for='formmake'>Make: </td> <td><select name="formmake" id="formmake" onchange="return populateModel(this.value);"> <?php $sql="SELECT id,manufacturer FROM manufacturer"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $data['id'] ?>" ><?php echo $data['manufacturer'] ?></option> <?php } ?> </select> </td> </tr> <tr> <td align="right" label for='formyear'>Year: </td> <td><select name="formyear"> <?php $sql="SELECT id,year FROM year"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $data['id'] ?>" ><?php echo $data['year'] ?></option> <?php } ?> </select></td> </tr> <tr> <td align="right" label for='formmodel'>Model: </td> <td> <select name="formmodel" id="formmodel"> <?php $sql="SELECT id,model FROM model WHERE year='1'"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $data['id'] ?>" ><?php echo $data['model'] ?></option> <?php } ?> </select> <!--<input name="formmodel" type="text" value="<?php echo $varmodel;?>" size="60" maxlength="100" />--> </td> I seem to be makeing a mess of this. I would like to thank everyone who looks at this for there help. Link to comment https://forums.phpfreaks.com/topic/265792-pass-information-from-selection-in-form-when-calling-from-db/ Share on other sites More sharing options...
alexmark Posted July 16, 2012 Author Share Posted July 16, 2012 Sorry I posted teh code backwards <tr> <td align="right" label for='formyear'>Year: </td> <td><select name="formyear"> <?php $sql="SELECT id,year FROM year"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $data['id'] ?>" ><?php echo $data['year'] ?></option> <?php } ?> </select></td> </tr> <tr> <td align="right" label for='formmake'>Make: </td> <td><select name="formmake" id="formmake" onchange="return populateModel(this.value);"> <?php $sql="SELECT id,manufacturer FROM manufacturer"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $data['id'] ?>" ><?php echo $data['manufacturer'] ?></option> <?php } ?> </select> </td> </tr> <tr> <td align="right" label for='formmodel'>Model: </td> <td> <select name="formmodel" id="formmodel"> <?php $sql="SELECT id,model FROM model WHERE year='1'"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $data['id'] ?>" ><?php echo $data['model'] ?></option> <?php } ?> </select> Link to comment https://forums.phpfreaks.com/topic/265792-pass-information-from-selection-in-form-when-calling-from-db/#findComment-1361995 Share on other sites More sharing options...
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